We set
\(u(t)&=\sum _{n=1}^{m}\langle u_0, \varphi _{n}\rangle E_{\alpha , 1}\left(-\lambda _{n} t^{\alpha }\right) \varphi _{n} + \sum _{n=m+1}^{\infty }\langle u_0, \varphi _{n}\rangle E_{\alpha , 1}\left(-\lambda _{n} t^{\alpha }\right) \varphi _{n} \\& =: u_1(t) + u_2(t).\) 

Then
\(\Vert u(0)\Vert ^{2}&=\sum _{n=1}^{m}\langle u_0, \varphi _{n}\rangle ^{2} + \sum _{n=m+1}^{\infty }\langle u_0, \varphi _{n}\rangle ^{2} \\& =: \Vert u_1(0)\Vert ^{2} + \Vert u_2(0)\Vert ^{2}.\) 

First, since \(\lambda _n\le 0\)  for \(n\le m\) , the function \(t \mapsto E_{\alpha , 1} (-\lambda _{n} t^{\alpha })\)  is non-decreasing for \(n\le m\) . Then \(E_{\alpha , 1} (-\lambda _{n} t^{\alpha }) \ge 1\)  for \(n\le m\)  and \(\Vert u_1(0)\Vert ^{2} \le \Vert u_1(T)\Vert ^{2} \le \Vert u(T)\Vert ^{2}\) . On the other hand, by using the
Hölder inequality we obtain
\(\Vert u_2(0)\Vert ^{2}&= \sum _{n=m+1}^{\infty }\langle u_0, \varphi _{n}\rangle ^{2} \\& = \sum _{n=m+1}^{\infty } \frac{\langle u(T), \varphi _{n}\rangle ^{2}}{\left(E_{\alpha , 1}\left(-\lambda _{n} T^{\alpha }\right)\right)^2}\\& = \sum _{n=m+1}^{\infty } \frac{\langle u(T), \varphi _{n}\rangle ^{2(1-\beta )}}{\left(E_{\alpha , 1}\left(-\lambda _{n} T^{\alpha }\right)\right)^{2}} \; \langle u(T), \varphi _{n}\rangle ^{2\beta }\\& \le \left(\sum _{n=m+1}^{\infty } \frac{\langle u(T), \varphi _{n}\rangle ^{2}}{\left(E_{\alpha , 1}\left(-\lambda _{n} T^{\alpha }\right)\right)^{\frac{2}{1-\beta }}}\right)^{1-\beta } \left(\sum _{n=m+1}^{\infty } \langle u(T), \varphi _{n}\rangle ^{2}\right)^\beta \\& \le \left(\sum _{n=m+1}^{\infty } \frac{\langle u_0, \varphi _{n}\rangle ^{2}}{\left(E_{\alpha , 1}\left(-\lambda _{n} T^{\alpha }\right)\right)^{\frac{2\beta }{1-\beta }}}\right)^{1-\beta } \Vert u(T)\Vert ^{2\beta }.\) 

Since \(-\lambda _{n} T^{\alpha } \le 0\)  for \(n\ge m+1\)  and \(0<\alpha <1\) ,
by Theorem 1.4 (p. 33-34) in [1]}, we see
\(E_{\alpha ,1}(-\lambda _nT^{\alpha }) = \frac{1}{\Gamma (1-\alpha )\lambda _nT^{\alpha }}+ O\left( \frac{1}{\lambda _n^2}\right)\) 
for all large \(n \in \mathbb {N}\) . Consequently we can prove
\(\left(E_{\alpha , 1}\left(-\lambda _{n} T^{\alpha }\right)\right)^{-1} \le C_1 (1+\lambda _n T^\alpha )\) 
for some constant \(C_1>0\) . See also [2]}.
Setting \(C_2=C_1\left(\frac{1}{\lambda _{m+1}}+T^\alpha \right)\) , we obtain
\(\left(E_{\alpha , 1}\left(-\lambda _{n} T^{\alpha }\right)\right)^{-1}\le C_2\lambda _n\) 
for \(n \ge m+1\) . Therefore
\(\Vert u_2(0)\Vert ^{2}& \le C_2^{2\beta } \left(\sum _{n=m+1}^{\infty } \langle u_0, \varphi _{n}\rangle ^{2} \lambda _n^{\frac{2\beta }{1-\beta }}\right)^{1-\beta } \Vert u(T)\Vert ^{2\beta }\\& \le C_2^{2\beta } \Vert u_0\Vert _{D((\kappa -A)^\epsilon )}^{2(1-\beta )} \Vert u(T)\Vert ^{2\beta },\) 

since \(\epsilon =\frac{\beta }{1-\beta }\) . We assume without loss of generality that \(\Vert u(T)\Vert  \le 1\) . Hence
\(\Vert u(0)\Vert  & \le \Vert u_1(0)\Vert  + \Vert u_2(0)\Vert \\& \le \Vert u(T)\Vert  + C_2^{\beta } R^{1-\beta } \Vert u(T)\Vert ^{\beta }\\& \le \left(1+ C_2^{\beta } R^{1-\beta }\right) \Vert u(T)\Vert ^{\beta }.\) 
